March 11th, 2020 at 6:41:01 PM
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You have 10 light bulbs going. Each has an average life of 1 year. The life expectancy of each bulb follows an exponential distribution. What this means is the bulbs are never overdue to burn out. Regardless of how long a bulb has been burning, it is no closer to burning out. The average remaining time is always one year per bulb. It is the same principle as waiting for a royal flush in video poker -- no matter how long you've been waiting for one, you're no closer to getting one (assuming no change in strategy).

That said, what is the mean waiting time for the LAST bulb to burn out?

Usual rules:

That said, what is the mean waiting time for the LAST bulb to burn out?

Usual rules:

- Please don't just plop a URL to a solution elsewhere until a winner here has been declared.
- All those who have won a beer previously are asked to not post answers or solutions for 24 after this posting. Past winners who must chime in early, may PM me.
- Beer to the first satisfactory answer and solution, subject to rule 2.
- Please put answers and solutions in spoiler tags.

It's not whether you win or lose; it's whether or not you had a good bet.

March 12th, 2020 at 3:56:38 PM
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Tic Toc. There's still team to join the prestigious beer club. I have had correct answers by PM from those already on it. This puzzle is pretty easy, I think, so if you've ever wanted to gain some math geek points here at WoV, this is your shot.

It's not whether you win or lose; it's whether or not you had a good bet.

March 14th, 2020 at 10:20:11 AM
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Any chance of an answer to this soon? I can't seem to find a function of probability of any sort versus time that works.

March 14th, 2020 at 3:25:43 PM
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I know better than to attempt math/probability problems (better to be thought a fool than speak up and remove all doubt), but I might have had a light turn on, so:

1/365 chance of a bulb burning out on that day, but 10 bulbs to start

so, expected time to lose first bulb is 10/365 or 1/36.5 per day, or 36.5 days average time to lose first bulb.

9 bulbs left, 9/365 chance per day (365/9=40.5556) so 40.5556 days for next bulb (77.056 days total)

8/365 chance per day (365/8=45.625) 122.681 total

7/365=52.1429 122.681+52.1429=174.8239 days so far

6/365; 365/6=60.8333; 60.8333+174.8239= 235.6572

5/365; 365/5=73; 73+235.6572=308.6572

4/365; 365/4=91.25; 91.25+308.6572=399.9072

3/365; 365/3=121.6667; 121.6667+399.9072=521.5739

2/365; 365/2=182.5; 182.5+399.9072=582.4072

1/365; 365/1=365; 365+582.4072=947.4072 days expected until all 10 burn out. (2.5956 years)

March 14th, 2020 at 5:18:50 PM
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This one is wide open to members and non-members of the beer club. I would hate to have to post a solution to this fairly easy problem. Very basic probability and statistics.

It's not whether you win or lose; it's whether or not you had a good bet.

March 14th, 2020 at 5:38:37 PM
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Quote:WizardThis one is wide open to members and non-members of the beer club. I would hate to have to post a solution to this fairly easy problem. Very basic probability and statistics.

Thanks Wiz, for making "us simpletons*** " feel even worse.

Is the solution that "easy", that it will probably be an "aha" or "light-bulb" moment for most of us that didn't think of it in time?

***: I am a "statistics simpleton", yet I also consider myself "above average at probability".

March 14th, 2020 at 6:35:12 PM
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Quote:ksdjdjThanks Wiz, for making "us simpletons*** " feel even worse.

I'm sorry. That could have been worded better.

I will say one probably needs to know the basics of the exponential distribution to solve this one. If you don't, I would leave this one alone.

It's not whether you win or lose; it's whether or not you had a good bet.

March 14th, 2020 at 8:56:46 PM
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Quote:Wizard(snip) I will say one probably needs to know the basics of the exponential distribution to solve this one. If you don't, I would leave this one alone.

I will probably leave this one alone, but do we need to work out the standard deviation to solve this puzzle and if yes is it (see spoiler)

Does the standard deviation equal to the "average life" (mean)?